The formula for the radiation pressure P in n-dimensional space for a given internal energy density u is ##\frac{u}{n}##.

I would really appreciate it if someone could provide a link that gives a simple derivation of this formula for dummies like me.

Here's my (possibly bogus) derivation: Consider a n-dimensional cube of sides [itex]L[/itex]. Its "volume" is [itex]L^n[/itex]. Its "area" is [itex]2 n L^{n-1}[/itex]. (Why is that? Well, a one-dimensional version of a cube is just a line. Its border consists of 2 points. A two-dimensional cube is just a square. Its border consists of 4 lines. A three-dimensional cube is a regular cube. Its border consists of 6 squares. The general case is that a n-dimensional cube has a border consisting of 2 x n faces, of which is a (n-1)-dimensional cube.)

So now imagine a single little packet of light at the center of an n-dimensional cube (with a "surface" made out of mirrors). It just bounces back and forth between the faces. Let's consider the case where the light is traveling in the x-direction. Then immediately before bouncing off the "face" that is perpendicular to the x-direction, its momentum in the x-direction is [itex]p = E/c[/itex] (because the energy-momentum relationship for electromagnetic energy is E = pc). Immediately after bouncing off, the momentum is [itex]p = -E/c[/itex]. The change in momentum is then [itex]-2E/c[/itex]. That means that the packet of light imparts a momentum change of [itex]+2E/c[/itex] to the mirror. Since the packet has to travel from one side of the cube to the other between bounces, and it travels at speed [itex]c[/itex], these bounces happen once every [itex]L/c[/itex] units of time. So the average force imparted on the mirrors by the light packet is:

[itex]F = \frac{\delta p}{\delta t} = \frac{2E/c}{L/c} = 2E/L[/itex]

Since pressure is force per unit area, you calculate the average pressure as the force [itex]F[/itex] over the area [itex]A[/itex]:

Another explanation is that the free electromagnetic field (as a massless vector field) doesn't contain any length or energy scale. That's why the action is invariant under scaling the coordinates and fields. The consequence is a conserved Noether currrent, and the conservation law just says that the covariant trace of the energy-momentum-stress tensor of the em. field vanishes: ##T_{\mu}^{\mu}=0##, but for a homogeneous situation (e.g., for thermal radiation) this means that ##u-nP=0##.

Here's my (possibly bogus) derivation: Consider a n-dimensional cube of sides [itex]L[/itex]. Its "volume" is [itex]L^n[/itex]. Its "area" is [itex]2 n L^{n-1}[/itex]. (Why is that? Well, a one-dimensional version of a cube is just a line. Its border consists of 2 points. A two-dimensional cube is just a square. Its border consists of 4 lines. A three-dimensional cube is a regular cube. Its border consists of 6 squares. The general case is that a n-dimensional cube has a border consisting of 2 x n faces, of which is a (n-1)-dimensional cube.)

So now imagine a single little packet of light at the center of an n-dimensional cube (with a "surface" made out of mirrors). It just bounces back and forth between the faces. Let's consider the case where the light is traveling in the x-direction. Then immediately before bouncing off the "face" that is perpendicular to the x-direction, its momentum in the x-direction is [itex]p = E/c[/itex] (because the energy-momentum relationship for electromagnetic energy is E = pc). Immediately after bouncing off, the momentum is [itex]p = -E/c[/itex]. The change in momentum is then [itex]-2E/c[/itex]. That means that the packet of light imparts a momentum change of [itex]+2E/c[/itex] to the mirror. Since the packet has to travel from one side of the cube to the other between bounces, and it travels at speed [itex]c[/itex], these bounces happen once every [itex]L/c[/itex] units of time. So the average force imparted on the mirrors by the light packet is:

[itex]F = \frac{\delta p}{\delta t} = \frac{2E/c}{L/c} = 2E/L[/itex]

Since pressure is force per unit area, you calculate the average pressure as the force [itex]F[/itex] over the area [itex]A[/itex]:

The units of pressure will no longer be the same in higher dimensions though. Is that supposed to be the case?

I think that the analogous notion of "pressure" does have different units in different numbers of dimensions. For example, in 2-D, you have a square with particles inside bouncing off the sides of the square. It wouldn't really make sense to characterize the pressure of the particles hitting the sides of the square in terms of force per unit area--the relevant number is force per unit length of the boundary.

I think that the analogous notion of "pressure" does have different units in different numbers of dimensions. For example, in 2-D, you have a square with particles inside bouncing off the sides of the square. It wouldn't really make sense to characterize the pressure of the particles hitting the sides of the square in terms of force per unit area--the relevant number is force per unit length of the boundary.

Another explanation is that the free electromagnetic field (as a massless vector field) doesn't contain any length or energy scale. That's why the action is invariant under scaling the coordinates and fields. The consequence is a conserved Noether currrent, and the conservation law just says that the covariant trace of the energy-momentum-stress tensor of the em. field vanishes: ##T_{\mu}^{\mu}=0##, but for a homogeneous situation (e.g., for thermal radiation) this means that ##u-nP=0##.

I'm gonna ask a dummy question; does this method say anything about the units of u and P?

Another explanation is that the free electromagnetic field (as a massless vector field) doesn't contain any length or energy scale. That's why the action is invariant under scaling the coordinates and fields. The consequence is a conserved Noether currrent, and the conservation law just says that the covariant trace of the energy-momentum-stress tensor of the em. field vanishes: ##T_{\mu}^{\mu}=0##, but for a homogeneous situation (e.g., for thermal radiation) this means that ##u-nP=0##.

which I think is the same as what vanhees71 has said.

Steven's derivation relies on the E=pc relation from Special Relativity though.

Am I right to say that the altered units of P, u and V are still dimensionally (units-wise) compatible with the other thermodynamic quantities of T, S, U and Q in this derivation?

The thermodynamic equation [itex]dU = TdS - PdV[/itex] shows that pressure has units [itex]\frac{energy}{volume}[/itex]. Energy has units (mass)(distance)2(time)-2. Volume has units (distance)n for n-dimensional space. So that means that pressure has units: (mass)(distance)2-n(time)-2. For 3-D this means the units:

(mass)(distance)-1(time)-2

which is the same as (force)/(area) = [(mass) (distance) (time)-2] (distance)-2

The thermodynamic equation [itex]dU = TdS - PdV[/itex] shows that pressure has units [itex]\frac{energy}{volume}[/itex]. Energy has units (mass)(distance)2(time)-2. Volume has units (distance)n for n-dimensional space. So that means that pressure has units: (mass)(distance)2-n(time)-2. For 3-D this means the units:

(mass)(distance)-1(time)-2

which is the same as (force)/(area) = [(mass) (distance) (time)-2] (distance)-2

Sorry, I was asking if the units were compatible with T, S, U and Q for any number of dimensions.